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Chemical Science - Kinetics (cont.) - Lecture 33

Principles of Chemical Science/nVideo Lectures - Lecture 33/nTopics covered:
/nKinetics (cont.)/nInstructor:
/nProf. Catherine Drennan/nTranscript - Lecture 33/nWe are getting close./nWe are in t... | more...Principles of Chemical Science/nVideo Lectures - Lecture 33/nTopics covered:
/nKinetics (cont.)/nInstructor:
/nProf. Catherine Drennan/nTranscript - Lecture 33/nWe are getting close./nWe are in the last unit. We are doing kinetics. And we are going to talk about reaction mechanisms today. We had a little bit of an introduction to that already last lecture setting up to do this./nToday we are going to run through and talk about how you come up with reaction mechanism, how you evaluate a proposed mechanism in terms of the experimental data. This is, again, in Chapter 13. All right./nSuppose you have a particular reaction. You know that 2NO gas plus O2 gas go to 2NO2 gas. And someone has come up with some experimental data for you. And they figured out that the rate law for this reaction, there is observed rate constant, so K observed./nAnd in terms of the NO concentration, it is second order. And in terms of O2, it is first order. And so now you can come up with a reaction mechanism that fits that experimental data. And, as I mentioned last time, you never know for sure that your reaction mechanism is correct./nAll you can do is be consistent with the data. You can prove reaction mechanism is wrong. You can come up with a mechanism and show that that cannot possibly be the mechanism given the experimental data./nYou can prove things wrong, but you can never be sure that you have it right. The best you can do is consistent, but you can prove a lot of things wrong. And if you have a lot of mechanisms and you prove a lot of them wrong and there is only one that seems to hold then there is a good chance that probably is correct, but you never 100% because someone might do another experiment and discover something else./nBut, in the problem set, you will be saying, yes, this mechanism is consistent, or you will be writing a consistent mechanism, or you will be saying that mechanism cannot possibly be true. All right./nLet's look at that. First let's just consider what the overall order of this reaction is. What is the overall order if this is the rate law? Three. I am just running through these. We will practice some of these because these are good couple point questions on the final, and so you don't want to miss these easy points./nAll right. If the overall mechanism is three, the overall order is three, three things coming together at once to react. Is it likely that is it one step? No, it is not. What is it called if you have three things coming together in terms of molecularity? Termolecular./nAnd those are rare, as we talked about last time. As I said, if you are getting together three people, it is usually not true that they all arrive sort of ready to go at the same exact moment. Usually one person is late, if not more./nGetting three things to react together at the same time is challenging, and so those are rare. So, this probably has more than one step. Then we have to try to divide up this reaction into steps and come up with a mechanism, a two-step mechanism we will try for this./nLet me leave that up. All right. Let's draw a potential mechanism on the board. And then we will try to figure out, for that mechanism, what the rate law would be if that was the correct mechanism and see if it is consistent with the experimental data./nAgain, what was the experimental rate law is K observed. That is the observed rate constant. Second order in NO and first order in O2. We can write the reaction in two steps, and each step is an elementary reaction so it occurs exactly as written./nIn the first step, we will have two NOs coming together. And the forward rate constant will be k1 and the reverse rate constant will be k-1. And it will be forming an intermediate N2O2. In the second step of the reaction we will have oxygen come in and interact with the intermediate, N2O2, and form, with a rate constant k2, two NO2s./nAnd that will give us our overall reaction. Now we need to write rate laws for each step. And since each step is an elementary reaction, we can write the rate law using the stoichiometry of that step as written./nThe rate law for the forward reaction would be what? What do I put first? k1. And then? NO squared. What about the rate for the reverse reaction? k-1. Times what? N2O2. OK. Let's look at each of these./nFor the forward direction, what is the order? The overall order is two. And so in terms of molecularity what is it? Bimolecular. For the reverse reaction the order is what? One. And what is that called? Unimolecular./nOK. Let's look at Step 2. The rate for step two is what? k2. Times? It would be O2 and N2O2, the concentration of each. Again, we can write them exactly as written. Now we consider the rate of N2O2 formation, so the overall rate of NO2 formation from these two steps./nAnd we can just use second step where NO2 is being formed. And so we can write this is equal to that rate law times two. Let me rewrite the rate law. And the two comes from the fact that two NO2s are being formed./nBecause two NO2s are being formed, this concentration is increasing at a different rate than the concentrations of these are decreasing. The book is inconsistent with whether it puts a two all the time, so I am not going to take off if people forget to put the two in this./nBecause the book should be more consistent with that. But, if you do see a two, it will mean that you are forming two of the products in the reaction. All right. Now we have written a rate. We are not equal to our experimental rate here because there is an intermediate./nThis is an intermediate. And an intermediate cannot appear in the overall rate law for something. You are going to be spending a lot of time getting rid of intermediates. Let's now leave this here and try to get rid of our intermediate./nI am having trouble with papers today. Now let's solve for the intermediate. And we want to solve for this intermediate in terms of things that are not intermediates. We want to solve for it in terms of reactants or products or rate constants./nWe need to consider then how this intermediate N2O2 is being formed and how this intermediate is being consumed or decomposed. We want to figure out the net formation of the intermediate. And the net formation of the intermediate is going to equal the rate at which it is formed./nIt is being formed over here in the forward direction of the first step, so we will write that down. We consider how it is being formed. It is being formed with the rate k1 times the concentration of NO squared, so that is just the rate law for the forward direction of the first step./nIt is also being decomposed in the first step in the reverse direction. It is being decomposed with the rate of k-1 times the concentration of the intermediate. That is just our rate law for this reverse step of the first reaction./nAnd it is also being consumed. It is also being consumed in the second step with a rate law of k2 times the concentration of the intermediate times the concentration of O2. That is just the rate law that we wrote for this second step./nIt is being formed, decomposed and consumed. The net formation equals the rate at which it is formed minus the rate at which it decomposes minus the rate at which it gets consumed. And now we can use something called the steady-state approximation./nIn the steady-state approximation, all it is saying is things are in a steady state. That means that the net formation of this intermediate is equal to zero or, if it is in a steady state that the rate at which it is being formed is going to be equal to the rate at which it is being consumed, either by the reverse Step 1 or Step 2./nThis rate is equal to this rate. That means it is steady, so it is in the steady state. The net formation is zero. The rate at which it is being formed equals the two rates at which it is being consumed or decomposed, the rate at which it is decaying are equal to each other./nThat is the steady-state approximation, and it is a very good approximation. And so you can use this for all of the mechanisms that you will be writing. That allows us to set it equal to zero, or we can rearrange these terms so that these two are on one side and the rate at which it is formed is on the other side./nLet's do that. And I am going to take these two and put them together and also pull out the term for the concentration of the intermediate because I want to solve for that term. I am going to try to pull it out and solve for it./nIf we rearrange this then, we get that the concentration of our intermediate, you pull it out from those terms, so that leaves k-1 plus the other term k2 times O2. Now I have just pulled this out from these two, and I have moved these to the same side of the equation./nAnd so that is going to be equal to k1 times NO squared. This is the rate at which you are decaying your intermediate and this is the rate at which you are forming that intermediate. Now I can continue to solve for this./nThe concentration of the intermediate is going to be equal to k1[NO]2 over k-1 plus k2 times the concentration of O2. Here is what the intermediate then is equal to. And now I can take this term and go plug it back in./nAll right. We can take the concentration and plug it back into here, so that whole term that we just came up with. And we will put it back into that equation. Then, if we substitute, we get the rate of NO2 formation is going to be equal to two rate constant k1, rate constant k2 times NO squared times the concentration of O2 over k-1 plus k2 times the concentration of O2./nBut that is not consistent. We have NO squared and two O2 terms. Whereas, in our observed experimental law, there is only one term for O2. That is not consistent so something is wrong. This is not consistent so there must be fast and slow steps./nIf there are fast and slow steps then we can simplify this equation and make it like the experimental equation. Now we can go over here and think about what would happen if Step 1 was fast. It is a fast and it is a reversible step./nIt is already written that it is a reversible step, so it is reversible. And Step 2 might be slow. Now we can consider, if we have a fast first step and slow second step, what that would do to this equation and whether we could then rearrange this equation such that it would be equal to our experimental rate law./nWhen we say something is slow, we are going to introduce a new term. And that the slowest step in a reaction mechanism is called the rate determining step. And, for a rate determining step, you can assume that that step is governing the overall rate of the reaction./nIt is slow enough that the rates of the other reactions don't really matter. That is what rate determining means. And let me just give you an example of a rate determining step. After today's lecture, you will be able to do all of the problems on Problem Set 10./nThis is your last problem set in 5.111, and so it would be really great when that problem set is finished and turned in. And so all through the rest of this lecture thinking about I can go out right after class and finish that problem set because you are very excited./nAnd by the time it gets close to the end, you see the end of the handout coming up. Your books are sort of packed, you are ready to go, and then the minute it is finished you are out the door. And you are going to some place where you can sit down and finish Problem Set 10./nYou are moving really fast. You might even be running. You may be even jumping over people in the infinite corridor to get to the library as fast as you can. You get to the library and you look for a table, but everyone else from 5.111 is already there and all the tables are filled./nYou go to the first floor and the second floor and the third. You go to the basement and in the stakes. Every table is filled up with people doing Problem Set 10. Then you have to leave the library./nYou think Building 2, lots of small classrooms, they cannot all be in use. You run over to Building 2, but all those classrooms are filled with 5.111 students doing Problem Set 10. And so eventually you get to Building 4 and you find a free classroom./nThe minute, the second you find that free classroom your books are out of your bad. They are out. Your calculator. Your correct kind of approved calculator is right there and you are ready to go. It takes you maybe two seconds to get out [10 at 2:50?] and it takes you maybe two seconds to pull out all your books to be ready to go, but it takes you a half an hour to find the table to do that problem set./nOverall, half an hour plus four seconds is a half an hour. That was the rate determining step. In finishing Problem Set 10 was finding that table. What you can do for these as well, if we say there is a slow step, you can assume that that is rate determining./nAnd you don't have to worry about how fast those other steps are. It just depends on the rate of the slow step. That is governing the overall rate at which the reaction or the process happens. That is rate determining step./nWe can use that now. And we can use that and go back and simplify the expression that we solved for the intermediate. Here we were solving for this expression, and we weren't considering anything about fast and slow steps at this point./nAnd so this is the equation that you would get if there were no fast and slow steps. And then we plugged it into this and found that it was not consistent. Now let's take a step back and go back here and think about what this expression would be if the first step was fast and the second step was slow./nHere, again, I have just put it up on the screen what we had on the board. We have this first step is fast and reversible and the second step is slow. Really, what we are talking about here is we are comparing rates for the decomposition of the intermediate, that is the reverse direction of step one, with the rate for the consumption of the intermediate./nWhich of these is faster? Well, decomposition is faster because this is a fast step. And the consumption in Step 2 is slow. What does that mean? Well, if we write it down then if the rate of the decomposition, this rate, the rate of the reverse reaction is faster then this k-1 is a bigger number than k2./nThe rate of consumption here is slow. We are going to consider the relative size of these numbers and how they fit in this term, and you can drop things out by assuming that one thing is a lot bigger than the other thing./nIf we do this then we are saying, again, this is the same thing that was on the last slide, that the rate of the decomposition is faster than the rate of consumption. i.e., Step 2 is slow, Step 1 is fast./nWhat you are saying is that k-1 is a bigger number. That rate constant is a bigger value than the k2 term. And, in the bottom of the equation here, in the bottom of this equation for the intermediate, we have a term k-1 and we have the term k2 times the concentration of O2./nAnd now, if you are saying this about the different rates of this reaction, if you are saying this is fast and that is slow, this is a big number, that is a smaller number. If you have on the bottom here a really big number plus a much smaller number in comparison, this smaller number is not adding much to the bigger number./nAnd we can just cancel that term out. We can just say it is so much smaller than the other, we are not going to worry about it, and we are going to get rid of it. Now our expression would look like this./nThe concentration of the intermediate is k1 times NO squared over k-1. We can rearrange this and think about does this look familiar to any of you in terms of what we have seen before. We can rearrange it to look like this./nAnd what is this term? What is this all equal to? K, the equilibrium constant. It is equal to k1. This is just an equilibrium expression. And, if you go back and look over here, you will see that if you are going to write the equilibrium expression for Step 1, products over reactants, that is what you would get./nAnd also, we saw last time, that you can express equilibrium constants in terms of rate constants. And the equilibrium constant for a reaction is going to be equal to the rate constant for the forward reaction over the rate constant for the reverse reaction./nAnd at equilibrium those rates are equal to each other. So this is just an equilibrium expression. When we assumed Step 2 was slow and Step 1 was fast, we can now solve for the intermediate in terms of an equilibrium expression./nLet's consider why this is true. OK. When you have a fast first step followed by the slow step, the first step is pretty much in equilibrium. The first reversible step is in equilibrium. And you can kind of think about it in terms of this diagram./nHere you have reactants going to intermediates, and this is fast and reversible. Then the second step, the consumption of the intermediate is slow going to products. If this is really pretty slow and not much of the intermediate is being siphoned off to product, more of it in fast equilibrium is going back and forth with the reactant, then that creates sort of an equilibrium expression./nIt is not perfect. Some of it is being siphoned off, but more or less you can consider this in equilibrium. Again, if you have a fast reversible step followed by a slow step, you can assume equilibrium conditions./nThat is going to make your life easier in terms of solving these expressions, because you can go right to writing an equilibrium expression then for your intermediate. Now we can go back and substitute./nWe have this term now for solving for our intermediate. Or, you could write rate constant k1 over k-1. Or, you could write big K, the equilibrium constant k1. These are equal to the concentration of the intermediate./nWe can put this term back in here now. This was, again, what we wrote in the very beginning for the formation of NO2. And we can substitute that back in. And we can have it in terms of the rate constants or using an equilibrium constant in there./nThose would be equivalent. And all of these Ks, in terms of the experiment, are going to be grouped together as a K observed. You usually are not going to measure all the individual rate constants./nYou are just going to have some overall rate constant that you do measure, an observed rate constant. And so if you write then either of these in that form where all your Ks just get grouped into K observed then you see that that is consistent with the experimental data./nThat mechanism works. A mechanism works that has two steps, the first step being fast and reversible and the second step being slow. We don't know for sure that that is the mechanism, but it is consistent with the experimental data./nLet's look at another example now. We are going to look at three different examples. Here there is a proposed mechanism that is very similar to the one that we just looked at. Now we are going to try to do this problem in a more efficient way than we did last time, which is kind of work through it the way you will when you are doing Problem Set 10 really fast./nAll right. We have a two-step mechanism, one that is fast and reversible and a step that is slow. The first thing you want to do is write the rate laws for each of the individual steps. For this one, what is the rate law for the forward reaction? k1 times O3./nThat is for the forward reaction. For the reverse reaction we have k-1 times O2 times the concentration of this intermediate O. Let's look at the second one. Here we have the rate k2 times the intermediate O times O3./nNow the rate is going to be determined by this slow step. And we can write the equation for the formation of O2 from this last step. And we could do that whether it was the slow step or not. We did that before as well./nIf we write the formation of O2, it is being formed in the last step from O and O3. We are using this rate law. And, again, we have a two in there because, like the last example, two things are being formed./nWe have to molecules of O2 being formed. The rate of formation of O2, two times k2 times this intermediate times O3. If there is no intermediate in the expression we would be done, but there pretty much always is going to be an intermediate in the expression./nWe need to solve for that intermediate. O is an intermediate, so you need to solve for it in terms of reactants, products and rate constants. Again, here we have this step that we saw before. The first step is fast and reversible, the second step is slow, and so you can assume, approximate that the first step is in equilibrium./nAnd so you can solve for the intermediate in terms of an equilibrium expression. What is the equilibrium expression for the first step? K would be equal to? I will just put it down here. You can write it in terms of k1 over k-1, products O2 times O over reactants O3./nThat is just our equilibrium expression. We have it in terms of the rate constants and in terms of this expression for products over reactants. And now we can take this and rearrange it to solve for our intermediate./nWe would have this, we would bring the O2 down to the other side, we would bring the O3 up here, which leaves us with k1 times the concentration of ozone over k-1 times the concentration of oxygen gas./nThat was a lot simpler than what we did over here. And so you can feel free, if you are told there is a slow step and a fast step, just jump right to that and solve it in terms of the equilibrium expression./nIt will save you some time. You should come up with the same answer if you do it the way that we did it in the first step. And sometimes you will have to because you won't know about fast and slow steps./nNow we can take this term and substitute it back into the rate law for the formation of O2 gas. And so we had two k2 times this intermediate times O3. We put this term in for our intermediate and we get two times k2 times k1 ozone squared over k-1 times O2./nAnd so this would be the rate law for this particular mechanism. If that is the rate law for that mechanism, you could think about what are the orders in this and could you experimentally go back and look and say, OK, is that consistent with the mechanism? Let's take a look at what we would predict from this./nFirst we could also write this in terms of K observed and group all the K constants together for K observed, and that is times ozone to the two and O2. If we look at this, if we were going to design experiments for someone to do, what would we predict to sort of prove that that is, in fact, a correct mechanism? What is the order, based on this rate law, in terms of the concentration of O3? Two./nIf someone went into the lab and doubled the concentration of O3, what should they observe happen to the rate? They should observe four times the rate. What is the order in terms of O2? What is it? Minus one, right./nIf you doubled the concentration of O2, what should happen to the rate, keeping everything else the same? It should be half. This is the kind of thing that can be experimentally tested, whether it is consistent./nAnd what is the overall order then of this reaction? It would be one. And so if you doubled both things, what should you observe happen to the rate? Yeah, it should double. You could go back and sort of test this./nOK. Let's look at one more example. This time we are not going to make a proposal ahead of time about fast and slow steps, but we are going to work out what the overall rate law for this is without fast and slow steps and then go back and consider which step, or any, are slow./nNow we are going to go through the long way and do this problem and then think about what we can learn from it in terms of we are given an experimental rate law. And once we see how this works out, we can think about whether a step is slow or not./nFirst we start the same way, and we are going to write the rate laws for each of the individual steps. Again, they are elementary reactions. Steps in a mechanism are elementary reactions. We can write the rate law exactly as the equation is written./nWhat is the rate law for the forward direction here of Step 1? First we have k1 times NO concentration times BR2 concentration. That is the forward. And in the reverse we have rate constant k-1 times NOBR2./nFor step two, we have rate constant k2 times NO. There you go. So, you know how to do these. We have that. Now we are going to be looking for the overall formation of NOBR, and so we can write this just using the last step./nWe can use that rate expression for that last step. The formation of this. And we are putting a two in because, once again, there are two of these being formed in that last step. Again, we could just use the rate expression for the last step./nAnd there is a two in there because two are being formed. That is the rate of formation of NOBR. Now we have an intermediate. This is being formed and is being consumed, so we need to solve for the intermediate./nThere cannot be any intermediates in our overall expression, so we need to get rid of that. So, it is intermediate and we need to solve for it. Now, we cannot go and just use the equilibrium expression for the top because we don't know anything about fast and slow steps./nWe don't know that that is going to be a good assumption. Let's just write it out the long way and then come back and consider whether that is consistent with the experiment or we have to do some modifications./nWe are going to solve for this in terms of reactants and products in Ks. Now we want to consider the net formation or the change in the concentration of this intermediate. The intermediate is being formed in the first step, so we can write down the rate expression for how it is formed, which is just this, k1 times NO times BR2./nThis is how it is being formed. It is decaying in this second step, in the backwards step, so we can put that in. We have a minus sign here. Because it is going away by the rate expression of k-1 times the concentration of the intermediate and it is also being consumed in the second step./nIt is being used up in that second step, so we will put that term in here. That is minus k2 times the concentration of the intermediate times NO. It is being formed in one step and it is being consumed in two different steps./nNow we can use the steady state approximation which says that the overall net change is zero. We can set this term equal to zero. It is a steady state approximation. That is the same as saying that the rate at which an intermediate is formed is equal to the rate at which the intermediate goes away./nWe can rearrange these terms, bring these on one side of the equation and have this one on the other. Rearranging then we moved these onto one side and the formation to the other side. Now we can pull out the term for our intermediate./nTake it out here, that leaves k-1. Take it out here, that leaves k2 and NO. Again, this is the rate at which it is being formed. And now we can solve for the intermediate. When we solve for the intermediate, we take this term and divide it by that term and we get this expression./nThat is solving for the intermediate. The next step is to take that intermediate and plug it back into the initial equation that we had, the formation of NOBR, and that was two times k2 times the intermediate concentration times NO./nNow we take this whole term and put it back into this term and we get this expression here. All right. This expression now is not consistent with this experimental rate law, so there must be fast and slow steps./nNow let's consider what is going to happen if the first step is slow and the second step is fast. Again, you are considering what is happening on the bottom of this expression. You are asking how does k-1 compare to this k2NO term./nAnd so if you say that the first step is slow then k-1 is small. Second step is fast. That means k2NO is a bigger term. If this second step is fast, the rate constant for the second step is a bigger number than the rate constant for the first step./nAnd so what you can do then is you can say, OK, that is small compared to this so we drop out that term. If we drop out that term, can we cancel anything else? What else can we cancel? We can cancel k2./nWhat else? We can cancel out one of the NOs. And that leaves us with this expression here. There is the expression again, two times k1NOBR2. And that can be written as K observed. And, look at that, it is consistent with the experimental rate./nNow let's just check the other way to make sure one should be right and one shouldn't be right. That we can get the answer if we go the other way around. This is consistent with the experiment. And the overall number of this is two./nAll right. Now let's look the other way around. If the first step is fast and the second step was slow that would mean that k-1 would be a lot bigger than the k2 term. k-1 rate constant for the first step./nk2 rate constant for the second step. Fast. Bigger than this. And so then we can take this term out. Can we cancel anything else? No, you cannot cancel anything else. This is the expression. We can put that down here./nWe can group the K terms to K observed. And so this is not consistent with the experiment that you had before. That way doesn't work. You can write out the whole thing and then think about whether they are fast and slow steps./nAgain, you will be asking the question which of these terms is bigger than the other term? What is the fast step? How does k-1 compare to k2? You can make the simplification. And only one should be consistent./nOne mechanism should be consistent. The other one should be inconsistent. OK. Now you can do Problem Set 10. | less...